Flutter/Dart 比较多个列表并生成包含所有给定列表中的元素的列表答案

【问题标题】：Flutter/Dart Compare multiple Lists and generate list with elements containd in all given listsFlutter/Dart 比较多个列表并生成包含所有给定列表中的元素的列表
【发布时间】：2021-01-23 14:09:39
【问题描述】：

所以这是我的问题：
假设我们有两个或更多列表

List<SomeModel> one = [SomeModel, SomeModel, SomeModel],
List<SomeModel> two = [SomeModel, SomeModel, SomeModel, SomeModel, SomeModel, SomeModel]

我想要做的是比较这两个/或更多列表并创建一个新列表
仅包含每个列表中存在的那些元素。

List<Fruits> one = [Banana, Apple, Cherry];
List<Fruits> two = [Cherry, Blueberry, Apple, Mango, Pineapple, Pear];
List<Fruits> three = [Cherry, Apple, Mango, Pineapple, Pear, Banana];

List<Fruits> getMatchingList(){
  do something...

  List<Fruits> matchingFruits = [Cherry, Apple];
  return matchingFruits
}

我真的不知道如何做到这一点，尤其是对于较大的列表或未知内容的列表。

【问题讨论】：

标签： flutter dart

【解决方案1】：

如果列表的数量是固定的，你可以使用 where 函数来过滤项目：

List<SomeModel> filteredList = one.where(item => two.contains(item))
     .where(item => three.contains(item)).toList();

所以你的方法是：

List<Fruits> getMatchingList(List<Fruits> one, List<Fruits> two, List<Fruits> three){
   return one.where(item => two.contains(item))
            .where(item => three.contains(item))
            .toList();
}

如果您需要处理越来越多的列表，另一方面，只需使用：

List<Fruits> getMatchingList(List<List<Fruits>> listsToAnalyze){
   Map<Fruits,integer> fruitCounter = new Map();
   //expand is a method that flattern a list of lists in a single list
   listsToAnalyze.expand(element => element).toList().forEach(element => {
    //if the element exists, its counter is increased, otherwise is set to 1
    fruitCounter.update(element, (count) => count+1, ()=> 1);
   }
   //remove results with counter < size of the list of lists
   //supposing that an item can appear only once in each list, this means
   //that the value appeared in all lists
   fruitCounter.removeWhere((key,value)=> value<listsToAnalyze.lenght);

   return fruitCounter.keys.toList();
   
}

如果一个项目可以在一个列表中出现多次，请不要展开列表并保留List<Fruit> 以控制在每个循环期间是否已在列表中插入项目。记得在每个新列表中清空地图。

List<Fruits> getMatchingList(List<List<Fruits>> listsToAnalyze){
       Map<Fruits,integer> fruitCounter = new Map();
       List<Fruits> fruitsInsertedThisRound= new List();
       //cycle on each list alone
       listsToAnalize.forEach(list => {
           fruitsInsertedThisRound.clear(); //ensure that the list is empty at each cycle
           list.forEach(el => {
           //if the element exists, its counter is increased, otherwise is set to 1
            if(!fruitsInsertedThisRound.contains(el){
              //fruit still not inserted this round
              fruitCounter.update(el, (count) => count+1, ()=> 1);
              fruitsInsertedThisRound.add(el); //add the element to the list of inserted fruits
            }
         }
       }

       //remove results with counter < size of the list of lists
       //this means that the value appeared in all lists
       fruitCounter.removeWhere((key,value)=> value<listsToAnalyze.lenght);

       return fruitCounter.keys.toList();
       
    }

【讨论】：

【解决方案2】：

通过删除重复合并两个列表

作为单行

a.addAll(b.takeWhile((item)=> !a.contains(item)));

通过循环

for(var item in a){
 if(!b.contains(item)){
  b.add(item);
 }
}

【讨论】：